Problem Description

You are given an m x n binary matrix mat (containing only 0s and 1s). Your task is to find the number of special positions in this matrix.

A position (i, j) is considered special if it meets the following conditions:

• The value at position (i, j) is 1 (i.e., mat[i][j] == 1)
• All other elements in row i are 0
• All other elements in column j are 0

In other words, a special position contains a 1 that is the only 1 in both its entire row and entire column.

The solution approach uses counting to efficiently solve this problem. It first counts how many 1s appear in each row and column using two arrays: rows and cols. The rows[i] stores the count of 1s in row i, and cols[j] stores the count of 1s in column j.

After counting, the algorithm traverses the matrix again. For each position that contains a 1, it checks if both rows[i] == 1 and cols[j] == 1. This condition ensures that the current 1 is the only 1 in its row and column, making it a special position. The answer counter is incremented for each special position found.

This two-pass approach has a time complexity of O(m × n) where m and n are the dimensions of the matrix.

Intuition

The key insight is that checking if a position is "special" requires verifying that it's the only 1 in its entire row and column. A naive approach would be to check every 1 we find by scanning its entire row and column, but this would be inefficient.

Instead, we can think about this problem differently: what if we knew in advance how many 1s exist in each row and column? If we had this information, determining if a position is special becomes trivial - we just need to check if the position contains a 1 AND its row has exactly one 1 AND its column has exactly one 1.

This leads us to a preprocessing step: we can make a single pass through the matrix to count the number of 1s in each row and column. We store these counts in two arrays - rows[i] tells us how many 1s are in row i, and cols[j] tells us how many 1s are in column j.

With this preprocessing done, we make a second pass through the matrix. For each cell that contains a 1, we check if rows[i] == 1 and cols[j] == 1. If both conditions are true, this 1 must be the only 1 in its row and column, making it special.

This counting strategy transforms what could be an O(m × n × (m + n)) problem (checking every 1 by scanning its row and column) into an O(m × n) problem (two passes through the matrix), making it much more efficient.

Solution Approach

The implementation follows a two-pass strategy using auxiliary arrays for counting:

Step 1: Initialize Counting Arrays We create two arrays to track the count of 1s:

• rows = [0] * len(mat) - stores the count of 1s in each row
• cols = [0] * len(mat[0]) - stores the count of 1s in each column

Step 2: First Pass - Count 1s in Each Row and Column We iterate through the entire matrix using nested loops:

for i, row in enumerate(mat):
    for j, x in enumerate(row):
        rows[i] += x
        cols[j] += x

Since the matrix is binary (contains only 0s and 1s), we can directly add the cell value x to our counters. When x = 1, it increments the respective row and column counters; when x = 0, it adds nothing.

Step 3: Second Pass - Count Special Positions We traverse the matrix again to identify special positions:

ans = 0
for i, row in enumerate(mat):
    for j, x in enumerate(row):
        ans += x == 1 and rows[i] == 1 and cols[j] == 1

For each cell, we check three conditions:

• x == 1: The current cell contains a 1
• rows[i] == 1: Row i has exactly one 1
• cols[j] == 1: Column j has exactly one 1

The expression x == 1 and rows[i] == 1 and cols[j] == 1 evaluates to True (which equals 1 in Python) when all conditions are met, and False (which equals 0) otherwise. By adding this boolean result to ans, we increment the counter only for special positions.

Time Complexity: O(m × n) where m and n are the dimensions of the matrix (two passes through the matrix)

Space Complexity: O(m + n) for the auxiliary arrays storing row and column counts

Example Walkthrough

Let's walk through the solution with a small example matrix:

mat = [[1, 0, 0],
       [0, 0, 1],
       [1, 0, 0]]

Step 1: Initialize Counting Arrays

• rows = [0, 0, 0] (one counter for each of the 3 rows)
• cols = [0, 0, 0] (one counter for each of the 3 columns)

Step 2: First Pass - Count 1s

We traverse each cell and update our counters:

• Position (0,0): value = 1 → rows[0] += 1, cols[0] += 1 → rows = [1, 0, 0], cols = [1, 0, 0]
• Position (0,1): value = 0 → no change
• Position (0,2): value = 0 → no change
• Position (1,0): value = 0 → no change
• Position (1,1): value = 0 → no change
• Position (1,2): value = 1 → rows[1] += 1, cols[2] += 1 → rows = [1, 1, 0], cols = [1, 0, 1]
• Position (2,0): value = 1 → rows[2] += 1, cols[0] += 1 → rows = [1, 1, 1], cols = [2, 0, 1]
• Position (2,1): value = 0 → no change
• Position (2,2): value = 0 → no change

After the first pass:

• rows = [1, 1, 1] (row 0 has one 1, row 1 has one 1, row 2 has one 1)
• cols = [2, 0, 1] (column 0 has two 1s, column 1 has zero 1s, column 2 has one 1)

Step 3: Second Pass - Find Special Positions

Now we check each cell that contains a 1:

• Position (0,0): value = 1, rows[0] = 1, cols[0] = 2 → NOT special (column 0 has 2 ones)
• Position (1,2): value = 1, rows[1] = 1, cols[2] = 1 → SPECIAL! (only 1 in its row and column)
• Position (2,0): value = 1, rows[2] = 1, cols[0] = 2 → NOT special (column 0 has 2 ones)

Result: 1 special position found at (1,2)

The position (1,2) is special because it's the only 1 in row 1 [0, 0, 1] and the only 1 in column 2 [0, 1, 0].

Solution Implementation

Time and Space Complexity

Time Complexity: O(m × n) where m is the number of rows and n is the number of columns in the matrix.

The algorithm consists of two passes through the entire matrix:

• First pass: Iterates through all m × n elements to calculate the sum of each row and column, taking O(m × n) time.
• Second pass: Iterates through all m × n elements again to check if each element meets the special criteria (element equals 1, row sum equals 1, column sum equals 1), taking O(m × n) time.

Total time complexity: O(m × n) + O(m × n) = O(m × n)

Space Complexity: O(m + n) where m is the number of rows and n is the number of columns.

The algorithm uses:

• rows array of size m to store the sum of each row: O(m) space
• cols array of size n to store the sum of each column: O(n) space
• A constant amount of additional variables (ans, loop indices): O(1) space

Total space complexity: O(m) + O(n) + O(1) = O(m + n)

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Condition Checking Order

A common mistake is checking if rows[i] == 1 and cols[j] == 1 without first verifying that mat[i][j] == 1. This can lead to unnecessary condition evaluations for cells containing 0, though it won't affect correctness.

Pitfall Example:

# Inefficient - checks row/column sums even for 0 cells
if rows[i] == 1 and cols[j] == 1 and mat[i][j] == 1:
    special_count += 1

Solution: Always check mat[i][j] == 1 first as a short-circuit condition:

# Efficient - short-circuits for 0 cells
if mat[i][j] == 1 and rows[i] == 1 and cols[j] == 1:
    special_count += 1

2. Attempting Single-Pass Solution

Some might try to identify special positions in a single pass while building the row/column sums, but this is impossible since you need complete row and column counts before determining if a position is special.

Pitfall Example:

# Wrong - trying to count special positions while building sums
for i in range(num_rows):
    for j in range(num_cols):
        if mat[i][j] == 1:
            row_sums[i] += 1
            col_sums[j] += 1
            # Can't determine if special yet - counts incomplete!
            if row_sums[i] == 1 and col_sums[j] == 1:
                special_count += 1  # This won't work correctly

Solution: Stick to the two-pass approach - complete all counting first, then identify special positions.

3. Matrix Dimension Assumptions

Assuming the matrix is square or non-empty without checking can cause index errors.

Pitfall Example:

# Dangerous - assumes matrix is non-empty
num_cols = len(mat[0])  # IndexError if mat is empty

Solution: Add edge case handling:

if not mat or not mat[0]:
    return 0
num_rows = len(mat)
num_cols = len(mat[0])

4. Misunderstanding "Special Position" Definition

Some might incorrectly think a position is special if it's the only 1 in either its row OR column (using OR instead of AND).

Pitfall Example:

# Wrong logic - uses OR instead of AND
if mat[i][j] == 1 and (rows[i] == 1 or cols[j] == 1):
    special_count += 1

Solution: Remember that a special position must be the only 1 in BOTH its row AND column:

if mat[i][j] == 1 and rows[i] == 1 and cols[j] == 1:
    special_count += 1